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Chemistry Fifth Edition Julia Burdge Lecture PowerPoints Chapter 3 Stoichiometry: Ratios of Combination 2020 McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. 3.1 Molecular and Formula Masses 2 Molecular and Formula Masses Although an ionic compound does not have a molecular mass, we can use its empirical formula to determine its formula mass.
2 2020 McGraw-Hill Education SAMPLE PROBLEM 3.1 Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: (a) propane, (C3H8). (b) lithium hydroxide, (LiOH). (c) barium acetate, [Ba(C2H3O2)2]. 3 2020 McGraw-Hill Education SAMPLE PROBLEM
3.1 Solution Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: (a) propane, (C3H8). (b) lithium hydroxide, (LiOH). (c) barium acetate, [Ba(C2H3O2)2]. Solution (d) 3(12.01 amu ) + 8(1.008 amu ) = 44.09 amu. (e) 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu. (f) 137.3 amu + 4(12.01 amu + 6(1.008 amu ) + 4(16.00 amu ) = 255.4 amu. 2020 McGraw-Hill Education 4
3.3 Chemical Equations 2 Interpreting and Writing Chemical Equations NH3 + HCl NH NH4Cl Ammonia and hydrogen chloride react to produce ammonium chloride. CaCO3 NH CaO + CO2 Calcium carbonate reacts to produce calcium oxide and carbon dioxide. 5 2020 McGraw-Hill Education 3.3
Chemical Equations 7 Balancing Chemical Equations Balancing a chemical equation requires something of a trial-and-error approach. In general, it will facilitate the balancing process if you do the following: 1. Change the coefficients of compounds (for example, CO2) before changing the coefficients of elements (for example, O2). 6 2020 McGraw-Hill Education 3.3
Chemical Equations 8 Balancing Chemical Equations 2. Treat polyatomic ions that appear on both sides of the equation (for example, CO32 as units, rather than counting their constituent atoms individually. 3. Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient. 7 2020 McGraw-Hill Education 3.4 The Mole and Molar Masses
2 The Mole A mole is a counting group , defined as the number of atoms in exactly 12 g of carbon-12. This number of atoms in 12 g of carbon-12 is known as Avogadros Number (NA): NA = 6.0221418 1023 objects If you order one dozen doughnut, you are asking for 12 doughnuts. If you order one mole of doughnuts, you are asking for 6.022 1023 doughnuts! 8 2020 McGraw-Hill Education 3.4
The Mole and Molar Masses 3 The Mole Conversion factors for moles of objects and number of objects: 23 1mol objects 6.022 10 objects 1 mol objects 6.022 1023 objects 23 6.022 10 objects 1 mol objects 9
2020 McGraw-Hill Education 3.4 The Mole and Molar Masses 4 Determining Molar Mass Chemists determine how many moles there are of a substance by measuring its mass (usually in grams). The molar mass of the substance is then used to convert from grams to moles. The molar mass (M) of a substance is the mass in grams of 1 mole of the substance. 10 2020 McGraw-Hill Education
3.4 The Mole and Molar Masses 5 Determining Molar Mass By definition, the mass of a mole of carbon-12 is exactly 12 g. Note that the molar mass of carbon is numerically equal to its atomic mass. Likewise, the atomic mass of calcium is 40.08 amu and its molar mass is 40.08 g, the atomic mass of sodium is 22.99 amu and its molar mass is 22.99 g, etc. 11 2020 McGraw-Hill Education
3.4 The Mole and Molar Masses 6 Determining Molar Mass 1 amu = 1.6611024 g NH 1g = 6.0221023 amu In effect, there is 1 mole of atomic mass units in 1 gram. So, the molar mass (in grams) of any compound is numerically equal to its molecular or formula mass (in amu). 12 2020 McGraw-Hill Education 3.4
The Mole and Molar Masses 7 Interconverting Mass, Moles, and Numbers of Particles Access the text alternative for these images 2020 McGraw-Hill Education 13 SAMPLE PROBLEM 3.6 Setup
Determine (a) the number of moles of C in 10.00 g of naturally occurring carbon and (b) the mass of 0.905 mole of sodium chloride. Setup The molar mass of carbon is 12.01 g/mol. The molar mass of a compound is numerically equal to its formula mass. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. 14 2020 McGraw-Hill Education SAMPLE PROBLEM 3.6 Solution
Solution 1 mol C 0.8326 mol C a 10.00 g C 12.01 g C 58.44 g NaCl 52.9 g NaCl b 0.905 mol NaCl 1mol NaCl 15 2020 McGraw-Hill Education SAMPLE PROBLEM 3.7
(a) Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water. (b) Determine the mass of 7.92 1019 carbon dioxide molecules. 16 2020 McGraw-Hill Education SAMPLE PROBLEM 3.7 Solution Solution 1mol H2O 6.022 1023 H2O molecules
a 3.26 g H2O 18.02 g H2O 1mol H2O 1.09 1023 H2O molecules 2 H atoms 1.09 10 H2O molecules 2.18 1023 H atoms 1H2O molecule 1O atom 23 1.09 10 H2O molecules 1.09 1023 O atoms 1H2O molecule 23 17 2020 McGraw-Hill Education
SAMPLE PROBLEM 3.7 Solution 2 Solution (b) 1mol CO2 44.01 g CO2 7.92 10 CO2 molecules 23 6.022 10 CO2 molecules 1 mol CO2 19 5.79 10 3 g CO2
18 2020 McGraw-Hill Education 3.4 The Mole and Molar Masses 8 Empirical Formula from Percent Composition With the concepts of the mole and molar mass, we can now use the experimentally determined percent composition to determine the empirical formula of a compound. 19 2020 McGraw-Hill Education
SAMPLE PROBLEM 3.8 Strategy Determine the empirical formula of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass. Strategy Assume a 100-g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles. 20 2020 McGraw-Hill Education
SAMPLE PROBLEM 3.8 Setup Setup The empirical formula of a compound consisting of N and O is NxOy. The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One hundred grams of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass contains 30.45 g N and 69.55 g O. 21 2020 McGraw-Hill Education
SAMPLE PROBLEM 3.8 Solution Solution 1 mol N 30.45 g N 2.173 mol N 14.01 g N 1 mol O 69.55 g O 4.347 mol O 16.00 g O N2.173O4.347 4.347/2.173 2
NO2 22 2020 McGraw-Hill Education 3.6 Calculations with Balanced Chemical Equations Topics Moles of Reactants and Products Mass of Reactants and Products 23 2020 McGraw-Hill Education 3.6
Calculations with Balanced Chemical Equations 4 Moles of Reactants and Products Balanced chemical equations give us the relative amounts of reactants and products in terms of moles. However, because we measure reactants and products in the laboratory by weighing them, most often such calculations start with mass rather than the number of moles. 24 2020 McGraw-Hill Education SAMPLE PROBLEM 3.11
Nitrous oxide (N2O) is commonly used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is NH4NO3 s N2O g 2H2O g (a) Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water produced in the reaction. 2020 McGraw-Hill Education 25 SAMPLE PROBLEM 3.11
Setup Setup The molar masses are as follows: 80.05 g/mol for NH4NO3, 44.02 g/mol for N2O, and 18.02 g/mol for H2O. 1 mol NH4NO3 1 mol N2O and 2 mol H2O 1 mol N2O 26 2020 McGraw-Hill Education
SAMPLE PROBLEM 3.11 Solution 1 Solution (a) 1 mol N2O 10.0 g N2O 0.227 mol N2O 44.02 g N2O 1 mol NH4NO3 0.227 mol N2O 0.227 mol NH4NO3 1 mol N2O 80.05 g NH4NO3
0.227 mol NH4NO3 18.2 g NH4NO3 1 mol NH4NO3 Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide. 27 2020 McGraw-Hill Education SAMPLE PROBLEM 3.11 Solution 2 Solution (b)
2 mol H2O 0.227 mol N2O 0.454 mol H2O 1 mol N2O 18.02 g H2O 0.454 mol H2O 8.18 g H2O 1 mol H2O Therefore, 8.18 g of water will also be produced in the reaction. 28 2020 McGraw-Hill Education 3.7 Limiting Reactants
Topics Determining the Limiting Reactant Reaction Yield Types of Chemical Reactions 29 2020 McGraw-Hill Education 3.7 Limiting Reactants 1 Determining the Limiting Reactant The reactant used up first in a reaction is called the limiting reactant, because the amount of this reactant limits the amount of product that can form.
When all the limiting reactant has been consumed, no more product can be formed. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. 30 2020 McGraw-Hill Education 3.7 Limiting Reactants 2 Determining the Limiting Reactant CO(g) + 2H2(g) NHCH3OH(l) Suppose that initially we have 5 moles of CO and 8 moles
of H2: 31 2020 McGraw-Hill Education 3.7 Limiting Reactants 3 Determining the Limiting Reactant 2 mol H2 moles of H2 5 mol CO 10 mol H2 1mol CO Because there are only 8 moles of H2 available, there is
insufficient H2 to react with all the CO. Therefore, H2 is the limiting reactant and CO is the excess reactant. 32 2020 McGraw-Hill Education 3.7 Limiting Reactants 4 Determining the Limiting Reactant To determine how much CO will be left over when the reaction is complete, we must first calculate the amount of CO that will react with all 8 moles of H2:
1 mol CO moles of CO 8 mol H2 4 mol CO 2 mol H2 Thus, there will be 4 moles of CO consumed and 1 mole (5 mol 4 mol) left over. 33 2020 McGraw-Hill Education 3.7 Limiting Reactants 5 Determining the Limiting Reactant When you use stoichiometry to calculate the amount of
product formed in a reaction, you are calculating the theoretical yield of the reaction. The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. It is the maximum obtainable yield, predicted by the balanced equation. 34 2020 McGraw-Hill Education 3.7 Limiting Reactants 6 Reaction Yield
In practice, the actual yieldthe amount of product actually obtained from a reactionis almost always less than the theoretical yield. The percent yield tells what percentage the actual yield is of the theoretical yield. actual yield % yield 100% theoretical yiel d 35 2020 McGraw-Hill Education
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